Question

An electrical motor spins at a constant 1975.0 rpm. If the armature radius is 7.112 cm, what is the acceleration of the edge of the rotor? O 305,200 m/s O 152.3 m/s O 15.20 m/s O 3042 m/s

Answer 1

Answer:

Option D is the correct answer.

Explanation:

Since value of angular acceleration is constant, the body has only centripetal acceleration.

Centripetal acceleration

               $a=\frac{v^2}{r}=\frac{(r\omega )^2}{r}=r\omega ^2$

We have radius = 7.112 cm = 0.07112 m

Frequency, f = 1975 rpm = 32.92 rps

Angular frequency, ω = 2πf = 2 x π x 32.92 = 206.82 rad/s

Substituting in centripetal acceleration equation,

              $a=r\omega ^2=0.07112\times 206.82^2=3042.17m/s^2$

Option D is the correct answer.